Question

The repeated factor of the determinant $\left|\begin{array}{ccc}y+z& x& y\\ z+x& z& x\\ x+y& y& z\end{array}\right|$

• A. $z-x$
• B. $x-y$
• C. $y-z$
• D. none of these

Answer 1

The correct option is A $z-x$

$\left|\begin{array}{ccc}y+z& x& y\\ z+x& z& x\\ x+y& y& z\end{array}\right|$
$R_1\to {R_1+R}_2+R_3$
$=\left|\begin{array}{ccc}2(x+y+z)& x+y+z& x+y+z\\ z+x& z& x\\ x+y& y& z\end{array}\right|$
$=(x+y+z)\left|\begin{array}{ccc}2& 1& 1\\ z+x& z& x\\ x+y& y& z\end{array}\right|$
$C_1\to C_1-2C_3,C_2\to C_2-C_3$
$=(x+y+z)\left|\begin{array}{ccc}0& 0& 1\\ z-x& z-x& x\\ x+y-2z& y-z& z\end{array}\right|$
$=(x+y+z)\left[\right(z-x\left)\right(y-z)-(z-x\left)\right(x+y-2z\left)\right]$
$=(x+y+z)(z-x{)}^2$
Therefore,$(z-x)$ is repeated factor of the given determinant.
Hence, option 'A' is correct.